Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,459 questions

17,854 answers

1,446 comments

807,730 users

Label any intercepts, relative extrema, points of inflection, and asymptotes.

0 votes
Analyze and sketch a graph of the function. Label any intercepts, relative extrema, points of inflection, and asymptotes. Use a graphing utility to verify your results.

y = (1/ x - 2) - 3
asked Jan 24, 2015 in CALCULUS by anonymous

2 Answers

0 votes

Step 1 :

The function .

Rewrite the function as

Graph the function .

image

 Domain :

 The function

The function continuous for all the points except at .

The domain of the function is .

Intercepts :

y - intercept is :

image

image

y - intercept is .

x - intercept :

Consider and solve for x.

x - intercept is .

Step 2 :

Asymptotes :

Vertical asymptote exist when denominator is zero.

Equate denominator to zero.

Vertical asymptote is .

Horizontal asymptote:

The line is called a horizontal asymptote of the curve if either

or  

image

image

Thus, the horizontal asymptote is .

answered Feb 19, 2015 by Sammi Mentor
0 votes

Contd...

Step 3 :

Intervals of increase or decrease :

Differentiate with respect to x:

image

is never zero in its domain.

f is decreasing on its domain because

Determination of extrema :

f is an decreasing function, hence there are no relative extrema.

Step 4 :

Differentiate with respect to x:

Determination of inflection point:

is never zero on its domain.

Hence, there is no inflection points.

Step 5:

Graph of the function  .

Plot the intercept points image and image

Draw the asymptotes and .

image

Verify the results by graphing utility.

Solution :

x - intercept is and y - intercept is .

The function does not have relative extreme.

The function  does not have inflection points.

The vertical asymptote is and  the horizontal asymptote is .

answered Feb 20, 2015 by Sammi Mentor

Related questions

asked Jul 23, 2014 in PRECALCULUS by anonymous
...